$ \left(\dfrac{1}{64}\right)^{-\frac{1}{3}}$
Answer: $= 64^{\frac{1}{3}}$ Figure out what goes in the blank: $\Big(? \Big)^{3}=64$ Figure out what goes in the blank: $\Big({4}\Big)^{3}=64$ So $\left(\dfrac{1}{64}\right)^{-\frac{1}{3}}=64^{\frac{1}{3}}=4$